Thanks for the kind words. Quadophile, that’s an interesting twist; here’s how I see it:
A typical dynamic driver is shaped like a cone with its top cut off--the large opening where it attaches to the speaker enclosure being the base of the cone and the area near the voice coil being the cut-off tip. A cone with its tip chopped off is called a "frustrum of cone." The volume of a frustrum of cone (i.e., the volume of air that will be moved as the diaphragm moves) is:
V = 1/3 * pi * h * [ r1^2 + r2^2 + r1*r2 ]
where pi is the universal constant 3.145962..., h is the height, as measured from the base to the top, r1 is the radius of the base, r2 is the radius of the top, and the notation "^2" means "squared;" i.e. r1^2 = r1*r1. When r2 = 0, this minimizes the volume to that of a true cone (V = 1/3 pi h r^2) and when r1 = r2 then it maximizes it as the volume of a cylinder (V = pi h r^2).
Notice that the volume increases with the square of the radii, so even small increases in the diameter of the driver can produce large increases in volume. Alternatively, you pay dearly with respect to volume when you reduce the diameter of driver. Also, even though r1 and r2 contribute identically algebraically to the equation, r1 is the radius of the opening, so r1 > r2, and therefore increases in r1 will deliver a greater increase in volume than the same linear increase in r2.
Now let's compare three drivers. The first is a 6.5" driver; actual drivers have an opening a little less than their nominal size, say 5.5" diameter, so the radius r1 is half this (r1 = 5.5/2). Let the radius at the voice coil end be r2 = 4.5/2 and for simplicity let the height h = 1. (A Parts Express catalog is a good source of numerous drivers and their measurements). So for our 6.5" driver:
V = 1/3 * pi * 1 * [ (5.5/2)^2 + (4.5)^2 + 5.5*4.5 ] = 13.5 cubic inches
By making r2 close in size to r1, we've really pushed the volume, since the volume is maximum when it is a cylinder. So, for an h = 1, we've been generous in getting a volume for the 6.5" driver.
Now let's compare that to two 12" woofers (both with r1 = 11"/2), one with a small radius near the voice coil (r2 = 2"/2) and one with a large radius over the voice coil (r2 = 6"/2). Real woofers would have a "height" (h) greater than 1", but let's leave h = 1 to focus the comparison on driver diameter size, which really is the important measurement for wide vs. narrow speaker cabinets. The two volumes are:
V = 1/3 * pi * 1 * [ (11/2)^2 + (2/2)^2 + 11*2 ] = 55.8 in^3
and
V = 1/3 * pi * 1 * [ (11/2)^2 + (6/2)^2 + 11*6 ] = 110 in^3
So these woofers have a diameter a little less than twice that of the 6.5" driver, yet move 4 - 8 times the volume of air--and in reality it would be more than that because they would likely have a h > 1. (Even just an h = 2 would double the multipliers to 8 and 16 times).
Now the 6.5" designer can increase the volume of air moved by increasing the driver's throw, or excursion. Maximum excursions are usually less than an inch, and often much less. Here's the rub: let's take a 1/2" excursion at 30 Hz. That means the diaphragm has to move 1/2" in 1/60 of a second (to max amplitude) and then back again in the next 1/60 of a second, so that the full wavelength is generated in 1/30 second. Moving 1/2" in 1/60 sec is an average of 1.7 mph. This may not sound like much, but it is an average rate and the diaphragm will actually have to accelerate up to something above that rate and then decelerate back to 0, reverse direction, accelerate, decelerate, etc. Now while all this is going on, the music is changing (since we rarely listen to a 30 Hz sine wave), so that the actual signal is a complicated supposition of amplitudes and frequencies, which arrive at the driver pretty much independent of where the diaphragm is in its excursion. Higher frequencies make the problem more difficult, since by definition the driver must produce a wave in less time, so the excursion distance is going to influence the high-end roll-off of the driver (where the driver just physically cannot move a significant amount of air oscillating at that frequency), and thus it affects the choice of mid-range, tweeter, etc. There is a real tussle and tuggle here for control of that diaphragm, and clearly, the longer the excursion, the more difficult it is to control.
So you either have to put a real juicer of an electromagnet in the driver to control the cone, or at some point I think speaker designers just say "Forget it!; I'll just add another driver" and double the volume of air being moved without changing the excursion. Two drivers seated at different points pumping at the same frequency just *have to* add complicated additive and subtractive nodes and troughs to the acoustic pressure, but somehow designers of speakers like Kef, VMPS, Hales, and many others, manage to do this and achieve excellent speakers. My guess is that your drivers are wired in parallel, so--if we hold reactance constant as an approximation--1/Z = 1/Z1 + 1/Z2, so two 8 Ohm drivers in parallel will look like a 4 Ohm load to the amp.
